\(\frac{-1}{39}+\frac{-1}{52}=\frac{-7}{156}\)

\(\frac{-6}{9}+\frac{12}{16}=\frac{-2}{3}+\frac{3}{4}=\frac{1}{12}\)

\(\frac{-2}{5}-\frac{-3}{11}-\frac{34}{37}-\frac{74}{-85}=\frac{-1215}{6919}\)

\(\frac{-5}{9}:\frac{-7}{18}=\frac{-5}{9}.\frac{-18}{7}=\frac{10}{7}\)

bấm mt

-1/39 + -1/52 = -7/156

-6/9 + -12/16 = -17/12

-2/5 - - 3/11 [ (-) + ( -) = +] = -2/5 + 3/11 = -7/55

-34/37 - 74/ -85 = -152/3145

-5/9 : -7/18 = 10/7

 -7/156

-17/12

-7/55

-152/3145

10/7

Quy đồng rồi tính là ra ngay mà

a: x-56:4=16

nên x-14=16

hay x=30

b: \(101+\left(36-4x\right)=105\)

\(\Leftrightarrow36-4x=4\)

hay x=8

Vì a: x-56:4=16

    =>x-14=16

   Hay:x=30

Vì b: 

   Hay:x=8

         Vậy x={30:8}

 Nếu sai thì thui nhá,đừng chửi mình!

\(\dfrac{5}{-6}=\dfrac{-55}{66};\dfrac{-10}{11}=\dfrac{-60}{66}\Rightarrow\dfrac{-55}{66}>\dfrac{-60}{66}\Rightarrow\dfrac{5}{-6}>\dfrac{-10}{11}\\ \dfrac{-3}{20}=\dfrac{-45}{300};\dfrac{2}{-15}=\dfrac{-40}{300}\Rightarrow\dfrac{-45}{300}< \dfrac{-40}{300}\Rightarrow\dfrac{-3}{20}< \dfrac{2}{-15}\\ -0,305>-0,36\)

\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)

Ta có \(:\)

\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)

\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)

Do \(55 < 60\)

\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)

Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)

\(\dfrac{-3}{20}\) và \(\dfrac{2}{-15}\)

Ta có :

\(\dfrac{-3}{20} = \dfrac{-3.3}{20.3}=\dfrac{-9}{60}\)

\(\dfrac{2}{-15} = \dfrac{2.4}{-15.4}= \dfrac{8}{-60}= \dfrac{-8}{60}\)

Do \(-9 <-8\)

\(=> \dfrac{-3}{20} < \dfrac{-2}{15}\)

Vậy \( \dfrac{-3}{20} < \dfrac{-2}{15}\)

\(A=\frac{-3}{5}+\left(\frac{-2}{5}+2\right)\)

\(A=\frac{-3}{5}+\frac{8}{5}\)

\(A=1\)

\(B=\frac{3}{7}+\left(\frac{-1}{5}+\frac{-3}{7}\right)\)

\(B=\frac{3}{7}+\frac{-1}{5}+\frac{-3}{7}\)

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\frac{-1}{5}\)

\(B=0+\frac{-1}{5}\)

\(B=\frac{-1}{5}\)

\(C=\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+\frac{52}{9}\)

\(C=\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\)

\(C=\frac{-7}{9}.1+\frac{52}{9}\)

\(C=\frac{-7}{9}+\frac{52}{9}\)

\(C=5\)

\(D=50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)

\(D=\frac{50}{100}.\frac{4}{3}.10.\frac{7}{35}.\frac{75}{100}\)

\(D=\frac{1}{2}.\frac{4}{3}.10.\frac{1}{5}.\frac{3}{4}\)

\(D=\left(\frac{1}{2}.\frac{1}{5}.10\right).\left(\frac{4}{3}.\frac{3}{4}\right)\)

\(D=1.1\)

\(D=1\)

A= 1

B= -1/5

C=5

D=1

mik có ý kiến lần sau bạn đặt câu hỏi nhớ ghi phân số trên dưới cho dễ nhìn nha

Các bạn làm chi tiết đc ko!!

f: \(-20:2^2+\left(-5\right)^9:5^8\)

\(=-20:4-5^9:5^8\)

=-5-5

=-10

g: \(-100:5^2+\left(-7\right)\cdot\left(-3\right)^2\)

\(=-\dfrac{100}{25}-7\cdot9\)

=-4-63

=-67

h: \(-84:4+3^9:\left(-3\right)^7+\left(-5\right)^0\)

\(=-21-3^9:3^7+1\)

\(=-21-3^2+1\)

=-20-9

=-29

i: \(-29-\left[16+3\left(51-49\right)\right]\)

\(=-29-16-3\cdot2\)

=-45-6

=-51

j: \(-58\cdot75+58\cdot\left(-50\right)-58\left(-25\right)\)

\(=58\cdot\left(-75\right)+58\left(-50\right)+58\cdot25\)

\(=58\left(-75-50+25\right)\)

\(=-58\cdot100=-5800\)

\(a,=75+60-13=122\\ b,=250-25+40=265\\ c,=4+100-9=95\\ d,=45+80-27=98\\ e,=5^3-100=125-100=25\\ g,=5+5=10\)

a)
=\(3.5.5+3.5.2-13=3.5\left(5+2\right)-13=15.7-13=105-13=92\)
b)
\(=5^3.2-25+2^3.5=5.25-5.5+8.5=5\left(25-5+8\right)=5.28=140\)
c)
\(6^2:9+50.2-3^3.3=36:9+100-3^4\)
=\(4+100-81\)=23

a: \(3\cdot5^2+15\cdot2^2-26:2\)

\(=75+60-13\)

=122

b: \(5^3\cdot2-100:4+2^3\cdot5\)

\(=250-25+40\)

=265

c: \(6^2:9+50\cdot2-3^3\cdot3\)

\(=4+100-9\)

=95

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...